IMPULSE AND MOMENTUM 239
where v0 and v are the velocities at the instants t = 0 and t = t, respectively. If v0 and v are parallel, equation (II) may be written in the form
L = mv—mvQ. (II')
192. Momentum.—The vector magnitude my is called momentum. Therefore the momentum of a particle equals the product of the mass by the velocity and has the same direction as the latter. Equation (II) states, therefore, that impulse equals the vector change in momentum.
PROBLEM.
Show that the component of the impulse along any direction equals the change in the component of the momentum along the same direction, that is,
f X dt = mx — mxQ, etc.
•/o
193. Dimensions and Units.—Substituting the dimensions of force and time in the definition of impulse and those of mass and velocity in the definition of momentum, we obtain [MLT-1] for the dimensions of both. The C.G.S. unit of
both impulse and momentum is the -—:-----. The British
^ sec.
unit is the pound-second.
Force and Momentum. — Let F denote the resultant of all the forces acting upon a particle of mass m. Then we have
F = m\f > '
-|(«v), (III)
which states that the resultant force experienced by a particle equals the time rate of change of the momentum of the particle. In order to extend this result to a system of particles let F denote the resultant of all the external forces acting upon the system. Further let Ft- be the resultant of all the forces acting upon any one of the particles. Evidently F* is the resultant of two sets of forces, namely, those which are